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The MKS unit of force is a Newton which is equal to a kg*m/s2.

Chapter 3 –

Force

Force and Newton’s Laws of Motion

Our study of kinematics (the description of motion) has given us the ability to determine how far, how fast, or how long an object moves. Now we will move to another branch of physics called dynamics, which is the study of the cause of motion. To better understand what causes motion, some 'ground' rules are necessary. These basic understandings are called Newton's Laws of Motion, and were stated by Sir Isaac Newton (1642-1727) as a synthesis and extension of others'

ideas. These laws explain motion in imaginary idealized systems (such as Physicsland) and deal with controlled situations. Only after understanding these three single situations is it then possible to investigate the causes of motion in our real world where forces cannot easily be isolated.

Situation: In a frictionless environment, a ball rolling along a level floor will continue move in a straight line at a constant speed until some net force (a push or pull) acts to stop it, slow it down, speed it up, or change its direction. A stationary object will remain at rest unless some unbalanced force is exerted upon it.

Situation: In a frictionless environment, a force acts on a wagon and causes it to accelerate. If the force is doubled, the acceleration of the wagon is doubled. If the original force acts on a wagon of twice the mass, the acceleration will be only half of the original acceleration.

Example #1: What force is needed to accelerate a 10-kg wagon at 3 m/s2, ignoring friction?

Given: m = 10 kg a = 3 m/s2

F = ? Equation: F = ma

Substitution: F = (10 kg)(3 m/s2)

Answer:F = 30 kg*m/s2 = 30N

(N is

Newtons)

Example #2: What is the acceleration of a 105-gram rocket in the first stage firing of its 2.7 N engine?

Given: m = 105 g = 0.105 kg F = 2.7 N

a = ? Equation: a = F/m

Substitution: a = 2.7 N/ 0.105 kg Answer: a = 26 m/s2

The force of gravity on an object can be found using Newton's Second Law. An object has "weight" only when it is under

Newton's Second Law of Motion:

If a net force acts upon an object, the object will accelerate in the direction of the force at a rate

proportional to the force, as well as inversely proportional to the object's mass.

Mathematically,

F = ma

, where F represents the

net force,

m represents

mass and a is acceleration

Newton's First Law of Motion:

Every object continues at rest or in a state of uniform motion unless acted upon by an

outside or unbalanced force.

Constants

Acceleration due to gravity

on earth = 9.8m/s2 or 32ft/

s2

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Chapter 3 –

Force

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Fakirs are Indian "magicians". They are supposed to have supernatural powers, and are able to do famous physical feats that defy the maxims of Western Science. For example how can Fakirs lie on a bed of nails without getting injured? The trick of the lying of nail has been long proven, as the number of nails helping. The greater number of nails ensured a lesser amount of pressure being inflicted on his skin, thus, the nails do not pierce through his skin.

Chapter 3 –

Force

Example #3: Find the weight of a box that has a mass of 25 kg.

Given: m = 25 kg ag = 9.8 m/s2

Fg = ?

Equation: Fg = mag

Substitution: Fg = (25 kg)(9.8 m/s2)

Answer: Fg = 245 kg*m/s2 250 N

The English unit for weight (or force) is the pound. Mass in the English system is measured in 'slugs'. To find your mass in slugs simply use the same equation, solved for mass, and English units.

Example #4: Find the mass of a woman who weighs 110 lbs.

Given: Fg = 110 lbs

ag = 32 ft/s2

Equation: m = Fg/ag

Substitution: m = 110 lbs 32 ft/s2

Answer: m = 3.4 slugs

Her mass is 3.4 slugs. What a beauty! Perhaps you can see why this term is not used very often. NOTE:

Situation: A foot kicking a ball exerts a force on the ball. From our experience we know that the ball also exerts a force on our foot since we can feel it, especially if we don't have any shoes on!

Newton's Third Law of Motion:

If an object exerts a force on another object, then the second object exerts a force on the

first equal in magnitude, but opposite in direction.

One kilogram equals 0.0685 slugs One pound equals 4.45 Newtons

Metric

Measurements

English

Measurements

Time

Seconds (s)

Seconds (s)

Distance

Meters (m)

Feet (ft)

Speed

m/s

ft/s

Acceleration

m/s

2

ft/s

2

Mass

Kilograms (kg)

slugs

Force

Newtons (N)

Pounds (lbs)

Question:

What units would weight have?

Answer:

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Chapter 3 –

Force

Example #5: Jason Elam's foot accelerates a 0.750-kg football from rest to 15 m/s in 0.42 second. Find the force of the football on Elam's foot.

The information given does not allow us to find the force of the football on Elam's foot directly. Instead we can use the information to find the acceleration of the football and then the force that Elam used to accelerate it. Once we know the force of Elam's foot on the football, we can use Newton's Third Law of Motion to determine that the force of the football on Elam's foot is the same magnitude, but in the opposite direction. A negative sign indicates that the force is in the opposite direction.

Given: m = 0.750 kg vi = 0

vf = 15 m/s

t = 0.42 s

Equation: F = ma (but first we need to find a)

a=

v

f

v

i

t

Substitution:

a

=

15

m

/

s

0

0 . 42

s

Answer: a = 36 m/s/s

Now we can substitute the calculated acceleration into F = ma to find the force of Elam's foot on the football. Substitution: F = (0.750 kg)(36 m/s/s)

Answer: F = 27 N

If the force of Elam's foot on the football is 27 N, then the force of the football on Elam's foot is the same magnitude, but in the opposite direction, -27 N.

Remember from the last chapter that objects accelerate toward the ground because of gravity. Thus, it is logical that the acceleration of a falling object is called the acceleration due to gravity and has the symbol ag (said " a-sub-g"). For objects

near the surface of the earth, the value of ag is 9.8 m/s2. If an object were taken to a different planet, the value of ag would not

be 9.8 m/s2 any longer. Some planets, like Jupiter, have more gravity. Other planets, like Mars, have less gravity. All objects

have some gravity. Even the moon, as small as it is, has some gravity. In fact, the acceleration due to gravity on the moon is 1.6 m/s2.

Because 9.8 is so close to 10, the value of 10 m/s2 is sometimes used where only an approximate answer is needed. In

fact, even the value of 9.8 m/s2 is only an approximation. As a person moves about on the surface of the earth the value will

vary because the earth does not have a uniform density and because the surface of the earth is not perfectly spherical. For example, in Paris ag is 9.81 m/s2. In Washington, D.C., the value agis 9.80 m/s2. But in Denver, Colorado, the "mile high city",

the value of ag is only 9.78m/s2. Of course, all of these round off to 9.8 m/s2 which is close enough for most physics work

because we rarely have more than two significant figures.

Weight

(Fg) is the

Force

with which mass is

pulled on by gravity. To calculate weight use:

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Chapter 3 –

Force

Problems – Force #1

1. Use Newton's first law to explain the following;

a) When people are riding in a car and it stops suddenly they are thrown forward.

b) A car is going around a curve at a speed that is far too fast, and the passengers slide to the side of the car. c) A coin is sitting on a record when the record player is turned on. It stays on for a short time then is thrown off. 2. On your own paper make a table like you see below and complete it using Newton's second law.

Net Force

(N) Mass(kg) Resulting Acceleration(m/s2)

1.6 0.8

30.0 0.5

22.0 11.0

12.0 4.2

14.4 .027

1.6 x 1011 6.3 x 10-23

1.3 x 1010 2397.0

2.78 x 10-12 2.1 x 10-25

3. A rocket sled is a rocket attached to a car-like vehicle that moves very fast horizontally. It is propelled by a solid fuel rocket motor that develops 900,000 N of thrust and burns for 3.7 seconds. The mass of the entire rocket sled (including fuel) is 4,500 kg.

a) What is the average acceleration of the rocket sled? b) What is the maximum speed of the rocket?

4. A replica of the standard kilogram (m = 1.0000 kg) was sent from Paris, France, and is kept in a vault in Washington, D.C. In Paris, it was weighed very carefully. In Washington it has also been weighed very carefully. However, the weights are not the same. There was no damage to the kilogram during its trip. What was the weight in; (HINT: make sure you read the pages introducing this new topic)

a) Washington, D.C.? b) Paris

5. A 2.7-kilogram mass is weighed in Denver, Colorado. Then it is carefully transported to the Kennedy Space Center where it is launched on a small-unmanned probe to the moon. It arrives at the moon undamaged. Find;

a) its mass on the moon. b) its weight on the moon. c) its weight in Denver.

6. A dog is taking his master for a walk. The dog strains at the leash with a force of 56.5 N. The master is wearing frictionless roller blades (the only way to walk your dog!) and has a mass of 72 kg. What is the acceleration of the master?

7. Explain what weight, mass and force are in relationship to each other.

* 8. Suppose that you were in a locked room in space (weightless conditions) with nothing but a mass known to be exactly 1.0 kg, a meter stick, a spring scale with all the number markings worn off, a frictionless rock with unknown mass, and a stopwatch. Describe in detail how you would find the mass of the rock in kilograms.

Answers:

2) Read down each column; F = 242 N, 0.39 N; m = 60 kg, 2.9 kg, 2.5x1033 kg, 1.3x1013 kg; a = 2 m/s2, 5.4x106 m/s2.

3a) 200 m/ss, 3b) 740 m/s 4a) 9.80 N, 4b) 9.81 N

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Chapter 3 –

Force

Force Problems #2

Use 9.8 m/s2 (OR 32 ft/s2) for a

g. Remember SIGNIFICANT FIGURES! (1kg on Earth = 2.2046 lb)

1. How much tension (another term for force) must a rope withstand if it is used to accelerate a 1700-kg car at 0.50 m/s2? Ignore

friction in this problem.

2. Use the scale and determine your weight in Newtons. Calculate your mass in kg. 3. A force of 240 N is needed to lift a Biology book. What is the mass of the text?

4. What force is needed to accelerate an ice hockey player (95 kg) across a frictionless surface at a rate of 8.25 m/s2?

5. A force of 250 Newtons is applied to a Physics cart that has a mass of 585 grams. Calculate the acceleration of the cart ignoring any friction.

6. What force would be needed to accelerate a bicycle of mass 80 kg (including rider) at a rate of 1.25 m/s2? Ignore friction.

7. A net force of 41.6 N accelerates a frictionless object at 4.21 m/s2. What is the mass of the object?

8. How much force is required to accelerate a 4.0 gram object at 10,000 "g's" (say, in a centrifuge)? A "g" is equal to the acceleration of gravity (9.8 m/s2), two "g's" would be two times the value of the acceleration of gravity (19.6 m/s2).

9. A shoeless swimmer has a mass of 95,000 grams. What force would he exert to do a pull-up with his shoes on (mass of 1.2 kg)?

10. What is the weight of a 70.0-kg astronaut

a) on earth? b) on the moon?

c) on Venus (ag = 8.7 m/s2)?

11. A force of 25 N is used to pull a frictionless wagon, mass of 20.0 kg, across an ice-covered pond. The wagon was initially at rest.

a) What is the wagon's acceleration?

b) What is the speed of the wagon after 3.0 seconds? c) What distance has been covered in 3.0 seconds?

12. A physics cart with a rocket attached to it has a mass of 1150 grams. It is traveling 3.5 m/s when the second stage engine ignites with 1.9 N of thrust for 1.6 seconds.

a) What is the speed of the rocket cart after the second stage finishes firing? b) What distance is covered during the firing of the second stage?

13. What average force is needed to accelerate a 240 gram baseball from rest to a speed of 30 m/s in 0.08 s?

14. What average force is needed to accelerate a 6.50 gram bullet from rest to 500.0 m/s over a distance of 0.7000 m along the barrel of a rifle?

15. According to a simplified model of a mammalian heart, at each pulse approximately 20.0 grams of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 second. What is the average force exerted by the heart muscle?

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Chapter 3 –

Force

17. A 0.10 gram spider is descending on a strand, which supports it (and keeps it from falling freely) with a force of 5.6 X 10-4 N.

What is the acceleration of the spider, ignoring air resistance?

18. A force of 20 lb is used to accelerate a 5 slug box sliding on a frictionless surface. What is the acceleration of the box? 19. A 15 lb bowling ball is accelerated by a force of 3 lb. What is the speed of the ball after a push lasting 1.3 s?

20. A 75 kg boy on ice skates pushes with a force of 30.0 N against the hands of a petite 45 kg girl, also on ice skates. Ignoring any friction, and knowing that each maintains his/her balance, what is the speed of each after pushing against each other for 0.90 s?

21. What is the net acceleration of a falling 65-kg skydiver if air resistance exerts a force upward of 250 N?

22. A 0.145 kg baseball traveling at 35.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What is the average force applied by the ball on the glove? What is the average force applied by the glove on the ball? 23. A typical home-run ball will leave the bat at a speed of 80.0 m/s after a contact time with the bat of perhaps 5.0 X 10-4 s. If the

baseball has a mass of 0.14 kg (and was initially at rest) what was the average force exerted by the bat on the ball? 24. What average force is required to stop a 1000.0 kg car in 6.0 seconds if it is originally traveling at 90.0 km/hr?

25. A 2000 kg car is pushed at a constant speed by 3 football players. Each football player can push with a force of 1300 N. What would be the acceleration of the car if a fourth football player (equal in all respects) were added?

26. A filing cabinet is pulled across the floor with a constant velocity by attaching a spring scale to the cabinet. When this is done, the reading on the scale is 4,215 N. Now the cabinet is pulled harder so that it no longer travels at a constant speed, but it accelerates at a constant acceleration of 0.21 m/s2. The scale reads 5,425 N when this is done. What is the mass of the filing

cabinet?

27. While bungee jumping the other weekend, Mrs. Roding (with her backpack full of physics books) fell 50 m before the bungee rope started to exert a force that eventually slowed her down. (Whew!) Find the initial force of the bungee rope if Mrs. Roding's acceleration rate dropped from 9.8 m/s2 to -1.3 m/s2. Not to worry, the bungee rope did its job and Mrs. Roding

eventually stopped before she hit the ground. (On a good day Mrs. Roding with her backpack has a mass of 95 kg)

*28. A 2.0 kg purse is dropped from the top of the Leaning Tower of Pisa and falls 55 m before reaching the ground with a speed of 29 m/s. What was the average force of air resistance?

Answers

1) 850 N3) 24 kg 4) 780 N 5) 430 m/s/s 6) 100 N

7) 9.88 kg 8) 390 N 9) 940 N 10a) 690 N, 10b) 110 N, 10c) 610 N 11a) 1.3 m/s2, 11b) 3.8 m/s, 11c) 5.6 m 12a) 6.1 m/s,

12b) 7.7 m 13) 90 N 14) 1160 N 15) 0.020 N *16) 0.85 m

17) 5.6 m/s2 18) 4 ft/s2 19) 8.3 ft/s 20) boy, 0.36 m/s; girl, 0.60 m/s 21) 6.0 m/s2

22) 807 N, -807 N 23) 22,000 N 24) 4200 N 25) 0.67 m/s2

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620 N

430 N R = 950 N 1 cm = 100 N

40º

Chapter 3 –

Force

Force and Vectors

As you recall, vector quantities have both a magnitude and direction, and so far we have worked with displacement and velocity vectors. Force is also a vector quantity. You could imagine that when pushing or pulling something, the magnitude of the force used and the direction the force is applied are both important!

Situation: A hockey player does a slap shot to make a goal. The amount of force (the magnitude) applied will determine the amount of acceleration and the final speed of the puck, and the direction the force is applied will aim it toward the goal.

Since often more than one force act on an object, it is often required that we add force vectors to solve problems. The resultant of force vector addition is referred to as Fnet (Net Force), and it is Fnet that allows us to calculate our anet (net acceleration).

Usual vector rules apply! Use arrow-heads, a dashed line for the resultant, and where asked to give a direction, measure counterclockwise from 0ْ (east).

Example: Two physics students are pushing their 82 kg phun physics project into the hall. They push at a 40º angle to each other. One student pushes with 430 N and the other with 620N. If they push for 3.0 seconds, what is the speed their project is moving? SOLUTION—

 Draw the force vectors to scale and with the correct angle.

 Add the vectors, and draw the resultant.

 The magnitude of the resultant is the net force.

 Use the net force to solve for the acceleration. Given: Fnet = 950 N

m = 82 kg Equation: a = Fnet/m

Substitution: a = 950/82 = 11.58 m/s2

Answer:a = 12 m/s2  Use the acceleration to find vf.

Given: a = 11.58 m/s2

vi = 0

t = 3.0 sec Equation: vf = at + vi

Substitution: vf = (11.58)(3.0) + 0 = 34.74 m/s

Answer: vf = 35 m/s

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Chapter 3 –

Force

Force Problems #3

You will need to make a drawing for each problem. Use a dashed line to show your resultant and label the resultant with an R. You do not need to state the direction, as it will be shown in your diagram.

1. Two forces of 2.0 N each act upon a 12-kg mass. What is the resulting force if the angle between the two original vectors is 75 degrees?

2. Two women wish to pull down a tree by means of a rope fastened near the ground to the trunk of the tree. If they use two ropes, they will pull the tree to a point between them where it will not land on either one of them. They tie the ropes at a 60o

angle from each other. If they each pull with a force of 420 N, what is the net force on the tree?

3. A force of 3.4 N pulls at an angle of 47 degrees to a second force of 8.3 N. What is the resultant force on the object? 4. Three fleas are making a physics philm. Each is attached to the same dry ice puck. The first flea pulls north with 5.0 N, the

second pulls east with 3.0 N, and the third pulls 45 degrees south of west with a force of 8.0 N. What is the net force on the puck?

5. Suppose the fleas in problem #4 are pulling with different forces. The one pulling north pulls with 4.6 N, the one east with 5.0 N. Now the third flea is very clever and wants to pull with exactly the right amount of force and in the correct direction to cancel the effect of the other two fleas. What should be the magnitude and direction of the force that the third flea uses? 6. A 6.5-kg mass is acted upon by two forces of 3.6 N each.

a) If these two forces act at an angle of 90 degrees with respect to each other, what is the magnitude of the resulting acceleration?

b) If the two forces act in the same direction, what is the magnitude of the resulting acceleration?

7. A force of 3.0 N and a second force of 4.0 N act on a 9.0 kg mass. The two forces act at an angle of 60from each other. a) If the object starts from rest, how fast will it be moving after 3.0 seconds?

b) What direction will the object be moving?

8. Two worms try to drag a 10.0-gram clod of dirt out of the way. The worms pull with an angle of 30 with respect to each other. If the first worm pulls with a force of 3.0 x10-8 N and the second with a force of 5.0 x 10-8 N, what is the distance the

clod will travel after the worms have pulled for 5.0 seconds?

9. Three chemists are trying to move a lab table. They are not aware of the advantages of knowing physics (typical chemists!). One pulls north at 25 N, the second pulls at 60o with 75 N, and the third pulls west with 55 N. They are pulling on a 90.0-kg

lab table with frictionless feet.

a) What is the acceleration of the table?

b) What is the velocity of the table after 8.0 seconds? c) How far does the table move in 8.0 seconds?

*

10. The net compressive force acting on the patella as a result of the action of the quadriceps muscle upward and the patellar tendon downward (Ft), as shown in the drawing at the right, is 35 N. Take the angle between the

muscle group and tendon to be 160°, and suppose the leg is bent symmetrically so that Fm = Ft.

a) What is the magnitude of each of the two forces Fm and Ft?

b) What happens to the force as the knee is further bent? Support your answer.

Answers:

1) 3.2 N; 2) 730 N; 3) 11 N; 4) 2.8 N;

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Chapter 3 –

Force

Force Problems #4

1. A force of 3.0 N is exerted on an object and it speeds up at the rate of 1.5 m/s2. What is the mass of the object?

2. A hunter shoots a silver winged bay turtle that is flying at an altitude of 250 meters. The mass of the turtle is 5.0 kg. The average air resistance on the turtle as it falls is 20. N.

a) What is the net force on the turtle? b) What is the acceleration of the turtle?

c) What is the speed of the turtle after 4.0 seconds? d) How far from the ground is the turtle after 4.0 seconds?

3. A block of mass 2.3-kg is pulled on a frictionless table by a constant force of 6.6 N. The block is initially at rest. a) What is the acceleration of the block?

b) What is the speed of the block at 3.0 seconds after the force starts acting? c) How far does the block travel in 2.5 seconds?

4. A blue-breasted-bent-beaked-barbaric buzzard is sitting in a beautiful baobab tree. It attacks two wiley wombats. What a mistake! The wombats capture the buzzard. They attach two log chains (mass negligible) to the buzzard and pull him to their cave. Since they wobble a lot, they pull at different angles to one another. Willie wombat pulls at 180 degrees and Wookie

pulls at 130 degrees. Willie pulls with a force of 65 N and Wookie pulls with a force of 45 N. Bonzo the buzzard has a mass of 15 kg.

a) What is the net force (magnitude and direction) on the buzzard? b) What is the acceleration of Bonzo?

c) How far will Bonzo have been pulled after 6.0 seconds?

5. Sally, Sue, and Simba are at a skating rink. They spot a handsome young man whose goal in life is to be a Physicist. They all start pulling on this young man. Sally pulls at 180 degrees with a force of 120 N, Sue pulls at 90 degrees with a force of 75 N and Simba pulls at 115 degrees with a force of 55 N. The Physicist has a mass of 80.0 kg and is on frictionless skates.

a) In what direction will the phuture Physicist be pulled? b) What is the magnitude of the force on the phuture Physicist? c) What is the acceleration of the phuture Physicist?

d) If the girls pull for 4.2 seconds, what will the speed of the phuture Physicist be after this time?

6. A force of 8.2 N accelerates a blue wad of gum at 16 m/s2. The same amount of force accelerates a red wad of gum at 28 m/s2.

What acceleration would the force of 8.2 N give if both wads were stuck together?

*7. The cable supporting a 2125 kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it give the elevator without breaking?

Answers:

1) 2.0 kg;

2) 2a) 29 N, down 2b) 5.8 m/s2 2 c) 23 m/s 2 d) 200 m

3a) 2.9 m/s2 3b) 8.6 m/s 3 c) 9.1 m;

4a) 100 N @ 159o 4 b) 6.7 m/s2 4 c) 120 m;

5a) 136o 5 b) 190 N 5 c) 2.4 m/s2 5 d) 10. m/s

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Chapter 3 –

Force

More about Vectors!

When discussing forces the resultant is called the net force (Fn). Once the value

for the net force is found, Newton's Second Law equation can be used. Sometimes the sum of all forces is zero, and we say that the object is in equilibrium. The forces creating an equilibrium situation are balanced, but not necessarily equal. If an object is in

equilibrium, the net force equals zero, and therefore the acceleration must be equal to zero.

The dinosaur at the left is standing in an elevator. There are two forces acting on the dinosaur; the force of gravity pulling down on the dinosaur, and the force of the elevator floor pushing upward on the dinosaur. The forces on the dinosaur are balanced. The force of gravity pulling down on the dinosaur is equal to the force of the elevator floor pushing upward on the dinosaur. The dinosaur is in equilibrium. There is zero net force on the dinosaur, and so the dinosaur will not experience any acceleration. It will maintain its constant state of motion.

When the elevator pushes upward on the dinosaur with a greater force than gravity pulls on the dinosaur, the forces on the dinosaur are unbalanced. There is a net force on the dinosaur in the upward direction, and therefore, the dinosaur will accelerate in the upward direction. See the next example.

Example: While riding the elevator at the Empire State Building a serious physics dinosaur (mass = 550 kg) experiences an upward acceleration of 1.2 m/s2. What is the

force of the floor pushing upward on the dinosaur? (Two more ways of asking the same question: If the dinosaur were standing on a scale, what would be his apparent weight during this upward acceleration? - OR - What force does he exert against the floor during this acceleration?)

In addition to pushing upwards on the dinosaur to balance the force of gravity, the elevator floor is pushing upwards to accelerate the dinosaur. This unbalanced or net force can be added to the upwards push to balance gravity to find the total force of the floor on the dinosaur.

Given: m = 550 kg

ad = acceleration of the dinosaur = 1.2m/s2

ag = 9.8 m/s2

Fn = mad = (550 kg)(1.2 m/s2) = 660 N

Fg = mag = (550 kg)(9.8 m/s2) = 5390 N

660 N + 5390 N = 6050 N  6100 N

The floor of the elevator is pushing upwards with 6100 N of force. Only 5400 N is needed to balance the pull of gravity on the dinosaur (his weight), so the remaining 660 N is used to accelerate the dinosaur at 1.2 m/s2. If the dinosaur were standing on a

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Chapter 3 –

Force

Force Problems #5

1. Instructions for an Estes Rocket kit include a graph of thrust capabilities for several sizes of engines. The C-6 series has an average thrust of approximately 7.0 N for a period of 1.7 seconds. The rocket we plan to use has a mass of about 115 grams. a. Find the net force on the rocket as it is launched.

b. Find the average acceleration of the rocket. c. Find the maximum velocity of the rocket.

d. Find the distance the rocket rises during the firing of the engine. e. Predict the maximum height of the rocket (ignoring air resistance).

f. Predict the amount of time that the rocket falls (with no chute and no air resistance). g. Predict the maximum height of the rocket assuming an average air resistance of 1.0 N.

2. Mrs. Roding has a mass of 58 kg on a good day. On one particularly good day, she decides to ride the elevator to the top of the Sears Tower instead of climbing the 104 flights of stairs. In her never-ending quest to show how useful physics is, she stands on a scale in the elevator as it begins its ascent. The scale shows 609 N. What is the acceleration of the elevator?

3. Mr. Spock used to enjoy riding the Tower of Doom while growing up on the planet Vulcan. He realized that he felt heavier on the way up as the ride accelerated him to the top. One day he measured the distance from the top to the bottom at 60 meters. The time that it took the ride to accelerate him to the top (assume constant acceleration) was 12 seconds. Spock had a mass of 27 kg as a child. Gravity on Vulcan is only 8.7 m/s2.

a) What was the force of the ride up on Spock?

b) Suppose that air resistance on Vulcan exerts a force of 23 N on Spock as he accelerates downward on the Tower of Doom. What was his acceleration when falling?

c) How long did he fall?

d) If the brakes work to stop the ride only during the last 3 meters, find the force on Spock at the end of the ride.

4. In a test using an unrestrained dummy (no seat belt), a car moving at 18 m/s (40 mi/hr) collides with a wall. The dummy is thrown forward and his head hits the windshield. The time for the dummy's head to come to rest, roughly 0.01 s, is much less than the time for the car to come to rest, about 0.1 s. If just before his head hits, the dummy is 'moving' at 18 m/s and his head weighs about 89 N, find the average force exerted on the dummy's head during the collision.

5. A small child (mass = 22 kg) slides down a pole at a playground with an average acceleration of 3.4 m/s2. What retarding

(slowing) force does the friction of her hands and the pole exert on her?

6. A 15-gram bullet traveling at 150 m/s is stopped by a tree after traveling 6.0 cm through the wood. What force did the tree exert on the bullet in stopping it?

*7. A fisherman in a boat is using a “10-lb test” fishing line. This means that the line can exert a force of 45 N without breaking (1 lb = 4.45 N).

a) How heavy a fish can the fisherman land if he pulls the fish up vertically at constant speed? b) If he accelerates the fish upward at 2.0 m/s2, what maximum weight fish can he land? [45 N, 37 N]

Answers:

1a) 5.9 N 1b) 51 m/s2 1c. 87 m/s 1d) 74 m 1e) 460 m 1f) 9.7 s 1g) 210 m.

2) 0.71 m/s2 3a) 260 N 3b) 7.8 m/s2 3c) 3.9 s 3d) 4000 N

(13)

k = _

Ff

_

Fg

Please note that the coefficient of friction () has no units and the forces we are using are defined in slightly different ways for more complex interpretations of the physics of friction

Fs –the force needed to overcome inertia

Fg – the weight of the object being pulled

s = _

Fs

_

Fg

Please note that the coefficient of friction () has no units and the forces we are using are defined in slightly different ways for more complex interpretations of the physics of friction

Ff–the force needed to maintain a constant speed

Fg – the weight of the object being pulled

k = _

Ff

_

Fg

_

Ff

_

=

weight

Chapter 3 –

Force

Friction And The Coefficient Of Friction.

Generally speaking, if you push or pull on an object at rest, it will resist movement because of the frictional force. Then when the block starts to move, it takes less force to keep the block in motion at a constant speed. The maximum force used to start the block in motion is known as the static coefficient of friction, while the force required to keep it at a constant velocity is known as the kinetic coefficient of friction. The coefficients of friction are not based just on the moving object, rather they are the result of the interaction of both the moving object and the surface it is moving across. Thus, the coefficient of friction will change if either the moving object or surface it is moving across changes.

The coefficient of friction () is a number that is the ratio of the resistive force of friction (Ff) divided by the perpendicular

force or weight (Fg) of the moving object. It is represented by the equation:

You can determine the coefficient of friction through experiments, such as measuring the force required to overcome friction or measuring the angle at which an object will start to slide off an incline. There are also charts of common coefficients of friction available

Different types of coefficient

The static coefficient and kinetic coefficient are the only types of frictional coefficients with which we will concern ourselves.

Static coefficient (s)

Static frictional forces from the interlocking irregularities of two surfaces will prevent any relative motion until some limit is reached. That threshold of motion is characterized by the static coefficient of friction. The static coefficient of friction is typically larger than the kinetic coefficient of friction.

Static friction exists when objects are not moving. An object on a flat surface is at rest, for example. Gravity pulls it down while the surface holds it up. The friction between the box and the surface makes it difficult to slide the box. When a strong enough force is applied, the box will slide, despite the frictional force.

Kinetic coefficient (k)

Once you overcome static friction, kinetic friction is the force resisting motion. Thus, kinetic coefficient of friction characterizes the force restricting the movement of an object that is sliding on a relatively smooth, hard surface.

A common way to measure the force between objects is to use the weight of the object. An object's weight is the force it exerts on another object and is caused by gravity. The friction equation for an object sliding across a material on the ground can be rewritten as:

Once you know the weight of the object, you can use a scale to measure the force it takes to begin to move the object. For example, measure the weight of a book. Then use the scale to measure the force required to start the book sliding along

(14)

Chapter 3 –

Force

To measure the static coefficient of friction, you read the value of the highest force as the object starts to move. Doing the same experiment for sliding or kinetic friction, you will take your reading while the object is moving at a constant velocity

(15)

Givens:

Ff =14.0 N

Fg or weight = 40.0 N

k = _

Ff

_

Fg

=

__

14.0_N_

40.0 N

= 0.350

Givens:

=0.350

Fg or weight = 80.0 N

Ff

=

k

Fg

=

0.350 * 80.0 N = 28.0 N

Chapter 3 –

Force

Sample problem # 1: A smooth wooden block is placed on a tabletop. A force of 14.0 N is necessary to keep the block moving at a constant speed along the tabletop.

a. What is the kinetic coefficient of friction if the weight of the block is 40.0 N?

b. If another block is placed on the first block for a total weight of 80.0 N, and the coefficient of friction remains the same, what force will be needed to keep the block moving at a constant speed?

Problems

1. A box weighs

40.0 N and a force of 10 N is necessary to keep the box moving at a constant speed as it is pushed across the floor. What is the kinetic coefficient of friction?

2. A box weighs 40.0 N and is slid across a surface that has a kinetic coefficient of friction with the box of .025. What force will be needed to keep the box moving at a constant speed?

3. An object is being pulled with a force of 10.0 N at a constant speed across a surface that has a kinetic coefficient of friction with the object of .025. What is the weight of the object?

4. A smooth wooden block is placed on a tabletop; a force of 25 N is necessary to keep the block moving at a constant speed. a. What is the coefficient of friction if the mass of the block is 7.4 kg?

b. If the weight of the block is increased to 108 N, what force will be needed to keep the block moving at a constant speed? 5. A student with a mass 65.0 kg is being pulled across a grassy lawn at a constant speed of 2.00 m/s with a force of 178 N.

a. What is the coefficient of friction?

b. If friend who weighs 755 N jumps on the student, what force will be needed to keep the 2 students moving at a constant speed?

6. A filing cabinet has a mass of 96.0 kg. A chemistry student wants to move it across the room and she wants to know the coefficient of friction for the floor and the cabinet. She hooks a spring scale to the filing cabinet and pulls with a maximum force of 102 N to get the cabinet moving. Once it is moving, she notices she only needs to pull with a force of 85.0 N to keep it moving at a constant speed. Be a smart physics student and help her by calculating the static coefficient of friction and the kinetic coefficient of friction.

*7. Police investigators, examining the scene of an accident involving two cars, measure 72 m long skid marks of one of the cars, which came to a stop before colliding. The coefficient of kinetic friction between the rubber tires and the pavement is 0.80. What was the initial speed of the car?

Answers:

1.) 0.250 2.) 1.0 N 3.) 400N 4a.) 0.34 4b.) 37 N

5a.) 0.279 5b.) 389 N 6) s = 0.108 k =0.090

(16)

Chapter 3 –

Force

Review

Problems

Use ag = 9.8 m/s2 on earth and ag = 1.6 m/s2 on the moon.

1.

Add the following vectors, showing the resultant with a dashed line.

a. F1 = 250 N, F2 = 350 N angle between F1 and F2 = 130˚ (state the direction as the angle the resultant makes with F1).

b. F1 = 250 N north, F2 = 500 N 40˚ north of west, F3 = 300 N south (state direction in terms of degrees relative to north,

south, east, and west).

2.

Two dogs are pulling a 90. kg sled at an angle of 35˚ to each other. If “Killer” pulls with 200. N and “Bitsy” pulls with 50. N, a. What is the net force on the sled?

b. How fast will it be moving after 2.0 seconds? c. How far will it have gone in 3.0 seconds?

3.

If all forces on an object balance out, which of the following statements are true (list all that apply): a. it is in equilibrium b. it could be stationary c. the net force is zero

d. it could be moving with constant speed e. it could be accelerating f. in a vector addition diagram, R is not zero

4.

A dog has a weight of 300 N on earth. What is her weight on the moon?

5.

A swimmer wishes to cross a river that is flowing at 270 degrees traveling at a speed of 18 m/s. She heads at180 degrees moving at a speed of 12 m/s. What is her velocity as observed by someone on shore?

6.

Some football players are running a pass play; the receiver runs 10 yards north, then 5 yards 40˚ east of north, then 15 yards east. How far and in what direction must the quarterback throw the football so that it gets to where the receiver ends up?

7.

A skydiver (72 kg) is about to jump from an airplane flying 1002 m up.

a. Before he jumps, he imagines the worst: that his parachute doesn’t open and there is no air resistance. If this were the case, how long would it take for him to hit the ground?

b. Fortunately, he jumped and his chute opened. There was 450 N of air resistance acting on the parachuted skydiver. How long did it take him to hit the ground?

8.

Two mad scientists are pulling on the last vial of “potion”. One pulls east with a force of 210 N. The potion (which is 5.0 kg) is accelerating west at 2.0 m/s2. How much force must the other mad scientist be pulling with, and in what direction must he

be pulling?

*

9. A flatbed truck is carrying a heavy crate. The coefficient of static friction between the crate and the bed of the truck is 0.75. What is the maximum rate at which the driver can decelerate and still avoid having the crate slide against the cab of the truck?

© Fyzics Softext Inc., page 72

Answers:

1a) 280 N @ 84° 1b) 460 N @ 142°

2a) 240 N @ 5° from “Killer” 2b) 5.3 m/s 2c) 12 m

3) A B C D 4) 50 N 5) 22 m/s, @ 236o

(17)

b) c) a)

Chapter 3 –

Force

Force Review:

To receive points you must write down both what the question asked as well as the answer.

For each of the variables, choose the units that best fits. You may use some answers more than once and some not at all.

1) Velocity a. m/s due north

2) Distance b. N

3) Speed c. s

4) Mass d. m due north

5) Force e. kg

6) Weight f. m/s2

7) Time g. m/s

8) Acceleration h. km

9) Displacement i. hours

10) Which situations have an acceleration of 0 and why or why not.

a. a car is driven in a straight line at a constant speed of 65 mi/hr for 10 min. b. a car is driven around a circular track at a constant speed of 65 mi/hr for 10 min. c. a car sits at a stop light for 2 min.

d. a car starts out at a stop sign and reaches a speed of 30 mi/hr in 2 min.

11) Describe the difference between velocity and speed and then distance and displacement.

12) Here in Denver, Colorado, why does a Feather take longer to fall to the ground than a hammer?

13) What is the acceleration due to gravity on earth using the: a) metric system b) the English system

14) If the force applied to an object is doubled and the mass remains constant what happens to the acceleration

15) If the force applied to an object is constant and the mass is doubled what happens to the acceleration

16) If a dog pulls east with 210N and the boy walking him pulls west with 220N, what is the net force?

17) What is equilibrium?

18) If an object is in equilibrium can it be sitting still?

19) If an object is in equilibrium can it be moving at a constant speed?

20) If an object is in equilibrium can it be accelerating?

21) Add these Force vector diagrams together and find the net force. Hint: use a scale of 1cm = 1 N

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Time (s) V elo cit y (K m /h r)

0

.

20

100

.

30

50

70

Graph #1

Chapter 3 –

Force

25) A car makes a fast 90 degree left turn, which of Newton’s Laws explains why the people in the car will seem to move right?

Force Review:

To receive points you must write down both what the question asked as well as the answer.

26) Comes to a sudden stop, which of Newton’s Laws explains why the people in the car will want to continue to move forward?

27) If you are standing still with your weight pushing down onto the earth, which of Newton’s Laws explains why the net force will be zero?

28) What mathematical formula does Newton’s Second Law Give us?

29) A tennis ball launcher shoots a tennis ball upward to max height of 4.5 meters above the launcher. What is the initial velocity of the ball?

30) If a tennis ball launcher shoots a tennis ball upward at an initial velocity of 9.4 m/s and the ball moves 42 cm in the launcher what is the acceleration of the ball in the launcher?

31) If a tennis ball launcher accelerates a tennis ball upward at 105 m/s2

and the ball has a mass of 55g, what force does the launcher apply to the ball?

32) If your mass on earth is 65 kg, what is your mass on the moon?

33) If your mass on earth is 65 kg, what is your weight on the earth?

34) If your weight on earth is 450N, what is your mass?

35) If your weight on earth is 450N, what is your weight on the moon?

36) What is the acceleration of the car as given by the Graph #1?

37) An 65 kg sky diver starts falling from rest out of a plane. His parachute opens immediately, and exerts a force of 450 N upward. a) What is the weight of the sky diver?

b) What is the net force acting on the sky diver? c) What is the net acceleration of the sky diver?

38) A skier with a mass of 65 kg goes skiing down a run and accelerates from rest at 4.4 m/s2 for 15 seconds, until she plows into a snow

bank, and comes to a stop in only 0.88 seconds. Find the force that skier experiences as she stops. a) What is the final velocity of the skier before she enters the snow drift?

b) What is the acceleration of the skier in the snow drift? c) What is the net force on the skier in the snow drift?

39) Mr. Cassady pulls on a rope south with 45N, Mr. Moriarty pulls east with a force of 66 N, and Mr. Evans pulls north with a force of 62N. What is the net force?

© Fyzics Softext Inc., page 74

Answers:

1) a 2) h 3) g 4) e 5) b 6) b 7) c & I 8) f

9) d 10a) no acceleration 10b) accelerating 10c) no acceleration 10d) accelerating 11) speed =scalar velocity = vector distance = scalar displacement = vector

12) air resistance 13a) 9.8m/s2 13b) 32ft/s2 14) acceleration doubles

15) acceleration is cut in half 16) 10N west 17) where the net force is zero 18) yes 19) yes 20) no 21a) 1.5N 21b) 1.5N 21c) 0 N 22) you must overcome inertia 23) 0.2N 24) His force is supported by all of the points on the nails 25) Newton’s 1st Law

26) Newton’s 1st Law 27) Newton’s 3rd Law 28) F = ma

29) 9.4m/s 30) 105m/s2 31) 5.8N

32) 65kg 33) 637N 34) 46 kg 35) 73N 36) 1.2 km/hr/s 37a) 637N down 37b) 187N down 37c) 2.9m/s2 down

References

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